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In gauge field theory its standard form is $A^{+}=n_{\mu }{A}^{\mu }=0$ for a fixed null direction $n$. A light-cone gauge uses light-cone coordinates to fix a longitudinal variable, often choosing $x^{+}$ as the evolution parameter for a relativistic particle phase-space action. Related field gauges set selected minus components to zero when a field has the required gauge invariance. A massive field without that gauge freedom must instead eliminate dependent components through its equations of motion.

First-class reduction. A regular set of independent constraints is first class when their Poisson brackets vanish on the constraint surface, equivalently locally $\{{\phi }_i,{\phi }_j\}=C_{ij}{}^k{\phi }_k$. The coefficients can be functions on phase space. With the generator $G={\sum }_i{\xi }^i{\phi }_i$, an infinitesimal canonical gauge transformation isEach independent first-class condition removes one phase-space dimension, and quotienting its independent gauge orbit removes another. Thus the regular physical phase space hasThis regular first-class phase-space reduction requires independent constraints and gauge directions. The count can fail at singular strata or for reducible constraints; first-class closure alone does not guarantee independence. For a concrete counterexample to counting redundant equations, take two canonical pairs and ${\phi }_1=p_1$, ${\phi }_2=2p_1$. Both equations are first class, but there is only one independent condition and one gauge direction. The surviving pair $(q^2,p_2)$ has dimension two, whereas blindly substituting $N=n=2$ gives zero.

Oscillator symplectic form. The center-of-mass pair $(x^m,p_m)$ describes translations and total momentum. The nonzero Fourier coefficients ${\alpha }_k^m$ describe the standing-wave modes compatible with free-end Neumann boundary conditions; their reality condition is ${\alpha }_{-k}={\alpha }_k^{\ast }$, and ${\alpha }_0^m=\sqrt{2{\alpha }^{\prime }}{p}^m$ in the standard normalization. For $k>0$, regarding ${\alpha }_k^m$ as a complex coordinate makes its conjugate momentum $i{\alpha }_{-k,m}/k$. Equivalently write ${\alpha }_k=(q_k+i{s}_k)/\sqrt{2}$. Up to a total derivative, its kinetic term is $-q_k\cdot {\dot{s}}_k/k$, a real canonical form. Inverting this oscillator symplectic form of an open string givesThe zero oscillator commutes with nonzero oscillators but is not independent of $p$: $\{x^m,{\alpha }_0^n\}=\sqrt{2{\alpha }^{\prime }}{\eta }^{mn}$. Brackets between the center pair and independent nonzero oscillators vanish.

The quadratic constraints obeyFor example, the two terms in the bracket with $L_n=\frac{1}{2}{\sum }_j{\alpha }_j\cdot {\alpha }_{n-j}$ give equal contributions after relabelling $j$. Applying this identity to both factors of $L_m$ givesThis classical Virasoro constraint algebra has no central term and closes on the constraints, hence is first class. For $G={\sum }_n{\xi }_{-n}{L}_n$, the oscillator gauge transformation isReality is respected when ${\xi }_{-n}={\xi }_n^{\ast }$.

Light-cone reduction and mass. Choose light-cone coordinates $X^{\pm }=(X^0\pm X^{D-1})/\sqrt{2}$; a vector square is $-2{\alpha }^{+}{\alpha }^{-}+{\mathbf{\alpha }}^2$, with $D-2$ transverse components. On the proposed gauge slice ${\alpha }_{k\ne 0}^{+}=0$, the variation becomesFor ${\alpha }_0^{+}\ne 0$, every nonzero-mode gauge parameter has an invertible coefficient, so the conditions locally fix the corresponding gauge freedom. Globally this is the usual patch in which $X^{+}$ is an admissible worldsheet clock; it is not a claim about strings for which that coordinate has turning points. The zero-mode reparameterization is left over.

On that slice the nonzero Virasoro constraints are linear in the longitudinal oscillators:No nonzero longitudinal oscillator remains independent. The residual action isThis residual mass-shell action in light-cone string gauge displays the remaining zero-mode constraint. At the quantum level normal ordering replaces its oscillator term by $N-a$. One can further set $x^{+}$ equal to time and solve for the light-cone Hamiltonian $p^{-}=[{\mathbf{p}}^2+(N-a)/{\alpha }^{\prime }]/(2p^{+})$.

Canonical quantization gives the transverse relationsHere $\hslash =1$, ${\alpha }_{k>0}$ annihilate the oscillator vacuum, and ${\alpha }_{-k}$ create excitations. Define $a_k^I={\alpha }_k^I/\sqrt{k}$ for $k>0$. The string level operator isThus it counts oscillator number weighted by mode number, and has nonnegative integer eigenvalues. The residual mass-shell condition givesThe constant $a$ is the normal-ordering constant of a string, or intercept, not an extra classical tension. In the usual Lorentz-invariant critical bosonic string, level one carries the $D-2$ transverse polarizations of a massless vector. A massive vector would need $D-1$ polarizations; the longitudinal one is not present. Lorentz consistency therefore requires that this vector level be massless, giving $a=1$. Equivalently the regularized transverse zero-point value $a=(D-2)/24$ gives $D=26$. This critical-vector assumption in the string intercept argument concerns the usual critical quantum theory. In $D=3$ the lone transverse level-one polarization transforms trivially under the transverse rotation group; a massive scalar interpretation is not excluded by counting. Thus the stated masslessness conclusion is not a consequence of polarization counting for arbitrary $D$. An arbitrary intercept in a transverse oscillator model also need not define the usual covariant critical theory.

The full self-dual massless spectrum. In the closed-string sector, $N$ and $\tilde{N}$ are the independent nonnegative integer oscillator levels of the two chiral sectors. The integer $n$ quantizes center momentum around the circle, $p_{\mathrm{circle}}=n/R$, and $w$ counts how many times the string winds it. At the self-dual circle $R=\sqrt{{\alpha }^{\prime }}$, zero mass requiresBoth levels are nonnegative, so their sum is at most two. Exhausting these possibilities gives the massless spectrum at the bosonic self-dual circle:

The last family is easy to miss because the uncompactified ground state is a tachyon; its positive compact energy cancels that negative contribution at these charges. There are no other possibilities: at level sum one, $n^2+w^2=2$ forces both charges to be $\pm 1$; at sum zero, $n^2+w^2=4$ gives exactly the four listed pairs. With $d=D-2$ transverse oscillators the number of independent massless polarizations is $d^2+4d+4=D^2$, hence 676 in the critical $D=26$ bosonic theory. The circle oscillator is included among the $d$ components; it should not be discarded when interpreting the lower-dimensional scalar states.

Use the Minkowski metric with signature $(+,-)$, so the Sine-Gordon equation is ${\varphi }_{tt}-{\varphi }_{xx}+\sin \varphi =0$. The light-cone coordinates here satisfyIn particular, there is no extra factor of four with this coordinate normalization. The Bäcklund transformation requires $a\ne 0$, because one of its equations contains $a^{-1}$.

Put $F=(\psi +\varphi )/2$ and $G=(\psi -\varphi )/2$. In the parameter convention of this paper the Sine-Gordon Bäcklund transformation gives $G_{+}=a\sin F$ and $F_{-}=a^{-1}\sin G$. Differentiating, for a twice differentiable transformed field, givesSince $\psi =F+G$, the addition formula yields ${\psi }_{+-}=\sin (F+G)=\sin \psi$. Therefore the transformed field satisfies the same Sine-Gordon equation:Subtracting the two differentiated equations also gives ${\varphi }_{+-}=\sin \varphi$, showing the compatibility with the seed equation. This proves the unheaded request before the numbered parts, without assuming a particular soliton form.

Use $x^{\pm }=(x^0\pm x^{D-1})/\sqrt{2}$, transverse coordinates $x^i$, $i=1,\dots ,d=D-2$, and the Minkowski metricThe relativistic particle phase-space action becomesIn the light cone gauge $x^{+}=t$, solve the mass-shell condition for $p_{+}$, assuming $p_{-}\ne 0$. The reduced phase-space action is $\int dt({\dot{x}}^{-}{p}_{-}+{\dot{x}}^i{p}_i-H)$, withThe last equality selects the future-directed momentum sector and makes positivity transparent. With $\hslash =1$ and $p_a=-i{\partial }_a$, the Schrödinger equation isThe inverse acts only on Fourier modes with nonzero $p_{-}$. Multiplication by $2p_{-}$ gives $2{\partial }_{-}{\partial }_{+}\Psi =({\Delta }_{\perp }-{\mu }^2)\Psi$. ThereforeThe light-cone Hamiltonian thus gives the same Klein-Gordon equation as covariant quantization.

For the massive two-form field, take $\mu \ne 0$. Apply ${\partial }^n$ to its field equation. Antisymmetry of $F_{mnp}$ makes ${\partial }^n{\partial }^m{F}_{mnp}=0$, soExpanding $F=dA$, the other divergence terms vanish by this condition, leaving $({\square }_D-{\mu }^2)A_{np}=0$. The light-cone decomposition of a massive two-form makes its dependent components explicit. The divergence equation isTaking $n=-$ and $n=i$, respectively, givesThe $n=+$ equation follows from these expressions: the two terms containing ${\partial }_{+}{\partial }_i{A}_{-i}$ cancel and ${\partial }_i{\partial }_j{A}_{ji}=0$. Consequently $A_{-i}$ and $A_{ij}$ are independent, each satisfying the Klein-Gordon equation with mass $\mu$. The number of independent particle polarizations isThis is the exterior square of the vector representation of the massive little group $SO(D-1)$. In the analogous Proca equation, ${\partial }^m{A}_m=0$ determines $A_{+}$ from $A_{-}$ and $A_i$, leaving $D-1$ components. A massive field has no gauge freedom that would justify setting these longitudinal components to zero. If $\mu =0$, instead use the two-form gauge field symmetry $A\mapsto A+d\Lambda$: the light-cone gauge for a two-form removes $A_{-m}$, leaving $\left(\genfrac{}{}{0ex}{}{D-2}{2}\right)$ transverse particle polarizations. The massive and massless counts are different.

In the closed-string mode expansion, $x^m,p_m$ are center-of-mass canonical variables, while ${\alpha }_k^i,{\tilde{\alpha }}_k^i$ are independent left- and right-moving transverse string oscillators. Their complex conjugates are ${\alpha }_{-k}^i,{\tilde{\alpha }}_{-k}^i$. The two zero-mode Lagrange multipliers impose the remaining mass-shell condition and closed-string level matching. The string level operators areTheir quantum definitions use normal ordering. The symplectic terms in the phase-space action givewith all brackets between distinct sectors zero. The nonzero-index string oscillators obey ${\alpha }_n^{i†}={\alpha }_{-n}^i$ and similarly for the right-moving sector. Define the momentum-labelled oscillator vacuum byFor $k>0$, $a_k^i={\alpha }_k^i/\sqrt{k}$ has $[a_k^i,a_l^{j†}]={\delta }_{kl}{\delta }^{ij}$. HenceStarting with $N|0;p⟩=0$, a finite product with $r_{ki}$ creation operators of mode $k$ has eigenvalue ${\sum }_{k,i}k{r}_{ki}$. The Fock space is generated by these products; both level operators have nonnegative integer eigenvalues. This establishes the integer string oscillator level property. Subtracting their physical zero-mode constraints enforces $N=\tilde{N}$.

There is a distinction between the displayed classical zero modes and their quantum constraints. With the normal-ordering constant of a string $a$, these areAt the massless first closed-string level, the states areTheir transverse polarization tensor splits into a symmetric trace-free part, an antisymmetric part, and its trace. These are the graviton, Kalb–Ramond field, and dilaton, with respective particle polarization counts $d(d+1)/2-1$, $d(d-1)/2$, and one. They have the transverse little group representations of massless particles. In a Lorentz-consistent bosonic string theory, the first chiral level is a massless vector, not a massive vector with one missing physical polarization; the closed-string products are therefore massless. This fixes $a=1$. Equivalently, regularized transverse zero-point energy gives $a=(D-2)/24$, and Lorentz consistency fixes the critical dimension of the bosonic string $D=26$.

It follows that the bosonic string mass spectrum isThe ground state has $M_0^2=-8\pi T$ and is a tachyon; level one is massless; for $N\ge 2$ the mass is $M_N=\sqrt{8\pi T(N-1)}$. The masslessness claim uses the consistent quantum theory, rather than an unshifted reading of the classical $L_0$.

A massive two-form at closed-string level two is present. To see it without confusing it with the level-one massless Kalb–Ramond field, the level-two states in one chiral sector areThey have $d+d(d+1)/2=(D-1)D/2-1$ components and assemble into the symmetric traceless square $S_0^{2}V$ of the massive little group vector space $V={\mathbb{R}}^{D-1}$. The full closed-string level is $S_0^{2}V\otimes S_0^{2}V$. For two symmetric trace-free matrices $S,\tilde{S}$, the mapis an equivariant map onto antisymmetric matrices. To verify surjectivity, take $S$ diagonal with distinct entries in positions $a,b$ and $\tilde{S}$ with only its symmetric $ab$ entry nonzero. Their commutator gives the $ab$ antisymmetric basis element. Finite-dimensional representations of the compact little group are completely reducible, so this quotient representation is also a subrepresentation. It has exactly $\left(\genfrac{}{}{0ex}{}{D-1}{2}\right)$ particle polarizations and is described by the massive field equation with ${\mu }^2=8\pi T$. At $D=26$ this gives 300 particle polarizations, consisting in light-cone coordinates of 24 components $A_{-i}$ and 276 components $A_{ij}$.

Use the Minkowski metric $\mathrm{diag}(1,-1)$ and the angular field $u=\beta \varphi$. The Euler-Lagrange equation becomes $u_{tt}-u_{xx}+m^2\sin u=0$. For the dimensionless light-cone coordinatesthis is $u_{\xi \eta }=\sin u$. This normalization keeps the coupling in the relation between the physical field and its angle, rather than silently setting $\beta =1$.

A Bäcklund transformation is a system of first-order differential relations that maps a solution to another solution. One convention for the Sine-Gordon Bäcklund transformation uses a nonzero parameter $a$ and defines $v={\mathcal{B}}_a[u]$ byA compatible initial value or integration constant selects a particular transformed solution. Put $w=(v+u)/2$ and $d=(v-u)/2$. Differentiating gives $w_{\xi \eta }=\cos d\sin w$ and $d_{\xi \eta }=\cos w\sin d$. Their sum and difference yieldThus compatibility of the first-order relations contains the field equations for both fields, and the transformed physical field $v/\beta$ also solves Sine-Gordon theory.

The transformation provides a generating conservation law. On the branch close to the original field write $v=u+2d$. Its first equation isIt recursively determines a formal small-$a$ expansion of $d$ in local derivatives of $u$:At every subsequent order the new coefficient occurs linearly, so this recursion continues indefinitely. The second Bäcklund equation and the first imply the exact identityIndeed ${\partial }_{\eta }\cos d=-a\sin d\sin (u+d)$ and ${\partial }_{\xi }\cos (u+d)=-a^{-1}\sin d\sin (u+d)$. Comparing powers of $a$ therefore gives local conserved currents. If ${\partial }_{\eta }{P}_j={\partial }_{\xi }{Q}_j$, our coordinate convention givesFor localized fields approaching vacua at spatial infinity, the boundary flux vanishes and ${\mathcal{I}}_j$ is conserved. The formal expansion need not converge: each coefficient is a separately exact local conservation law.

For example $P_0=u_{\xi }^2/2$, $Q_0=1-\cos u$, a light-cone combination of energy and momentum. The next coefficient is a derivative improvement, so it contributes no independent charge under the same decay conditions. At order $a^2$, remove the improvement generated by ${\partial }_{\xi }(u_{\xi }{u}_{\xi \xi })$ and multiply by $-2$. One obtains the genuinely higher conservation lawIt can also be checked directly using $u_{\xi \eta }=\sin u$. Continuing the recursion, and using the opposite light-cone construction, produces the local conserved-charge hierarchy of sine-Gordon theory, with infinitely many nontrivial higher-spin charges after derivative improvements are removed. This is the Bäcklund generating current for sine-Gordon conserved charges. The hierarchy is the characteristic field-theory form of classical integrability; an ordinary energy conservation law alone would not supply these constraints.

The same transformation constructs solutions rather than only currents. Starting from the vacuum $u_0=0$, its two first-order equations integrate toFor $a=e^{\rho }>0$ the exponent is $m\cosh \rho x-m\sinh \rho t+{\delta }_a$. This is a Sine-Gordon kink with velocity $\tanh \rho$, center set by ${\delta }_a$ and classical rest mass $8m/{\beta }^2$. Negative parameters can supply the corresponding opposite-orientation seeds.

The allowed Bianchi permutability for sine-Gordon Bäcklund transformations gives the two-step field algebraically. For vacuum seed and $a\ne b$,Branches of the inverse tangent must be continued smoothly; its principal value alone does not specify the vacuum labels of a multi-kink field. All angles here are $u=\beta \varphi$. In the physical-field version each field difference in the superposition formula carries $\beta$; the formula printed without it implicitly uses the $\beta =1$ angular-field convention.

To display two real scattering solutions, take $0<v<1$, set $\Gamma =(1-v^2{)}^{-1/2}$, and put $X=m\Gamma x$, $Y=m\Gamma vt$. Choosing $a=e^{\rho }$, $b=e^{-\rho }$ with $v=\tanh \rho$ and zero phase constants yields the Sine-Gordon kink-antikink scattering solutionChoosing instead $b=-e^{-\rho }$ yields the Sine-Gordon two-kink solutionThe latter has net angular winding $4\pi$, while the former has zero net winding. At large positive or negative time they separate into localized kinks with velocities $\pm v$. Repeated commuting transformations give general multi-soliton fields. Continuing the kink-antikink velocity to an imaginary value also gives a real Sine-Gordon breather:up to the irrelevant overall field sign and translations.

The scattering interpretation ties the construction to the conserved hierarchy. In an exact multi-soliton sector, the incoming species and rapidities reappear after the collision, up to permutation: the solitons change positions, not their asymptotic shapes or velocities, and the collision emits no radiation. In the two-kink example the large-time centers obey $|X|=|Y|-\log v+o(1)$, so a right-moving trajectory acquires a shift $-2\log v/(m\Gamma )$. These shifts encode the interaction even though the collision is elastic. The commuting construction makes the net displacement in a many-soliton collision the sum of its pairwise displacements, independent of how the collisions are ordered; this is pairwise additivity of soliton shifts and the classical counterpart of factorized scattering.

Conservation of the entire hierarchy is much stronger than conservation of energy and momentum: its independent rapidity-weighted sums constrain the whole asymptotic soliton data and rule out particle production in this sector. Generic initial fields may also contain radiation scattering data, so this statement is about the exact soliton collisions, not a claim that every initial field is a pure soliton. The Bäcklund map unifies the conserved hierarchy, explicit soliton construction and elastic, pairwise scattering picture of classical integrability.

Solving the massive subsidiary conditions. In light-cone coordinates, use $d{s}^2=-2d{X}^{+}d{X}^{-}+{\delta }_{IJ}d{X}^{I}d{X}^J$ and write $d=D-2$. The divergence condition readsThus, when ${\partial }_{-}$ is invertible,First apply this with $n=-$, then with $n=I$, and finally with $n=+$; symmetry supplies the mixed components already determined. The trace condition becomesThe independent components are $h_{--}$, $h_{-I}$ and the trace-free part of $h_{IJ}$. Under transverse rotations they form a scalar representation, a vector representation, and a symmetric traceless rank-two tensor. ThereforeThis is the light-cone decomposition of a massive spin-two field for $D\ge 3$. In $D=2$, the trace and divergence give $h_{+-}=h_{++}=0$ and ${\partial }_{+}{h}_{--}=0$. The massive wave equation then forces $h_{--}=0$, so there are no polarizations, consistent with the zero value of the printed count. The massive particle little group is $SO(D-1)=SO(d+1)$, and its symmetric traceless square branches asThe mixed components with the extra direction give the vector; one independent trace combination gives the scalar. These are exactly the polarizations of a massive spin-two field. The remaining independent components retain the massive Klein-Gordon equation.

Transverse bosonic modes and mass levels. The variables ${\alpha }_k^I$ are Fourier amplitudes of the physical transverse open-string mode expansion. Classically, reality requires ${\alpha }_{-k}^I=({\alpha }_k^I{)}^{\ast }$. The symplectic term in the action fixes their quantum commutators:With string tension convention ${\alpha }^{\prime }=(2\pi T{)}^{-1}$, the zero mode of the constraint givesThe longitudinal nonzero modes have already been removed in light-cone gauge in string theory. Quantum normal ordering introduces the string intercept $a$, giving the open bosonic string mass spectrumEach bosonic occupation number is a nonnegative integer, so $N$ is a nonnegative integer weighted by oscillator frequency.

Suppressing the common momentum label, the lowest light-cone levels of an open bosonic string areThe oscillator vacuum is annihilated by every positive ${\alpha }_k^I$. At level one there are only $d$ vector polarizations. For a Lorentz-consistent vector, these are the transverse polarizations of a massless particle, transforming under the rotation part of its massless particle little group. A massive vector would need $D-1$ polarizations, including a scalar under $SO(d)$ that is absent here. Thus the first bosonic vector level must be massless, fixing $a=1$.

At level two the commuting creation operators give a symmetric square. Its scalar trace and symmetric traceless square, together with the mode-two vector, are the massive-spin-two decomposition above. In the consistent bosonic theory they form one massive spin-two field with $m^2=1/{\alpha }^{\prime }$. The covariant equations describe its propagation while eliminating the redundant components. For $D=26$ the transverse counts are $300+24=324$, the symmetric traceless rank-two tensor dimension of $SO(25)$.

Half-integer fermionic modes. The Neveu–Schwarz sector has antiperiodic worldsheet Majorana fermions. Its Neveu–Schwarz fermionic oscillators obeyThe oscillator vacuum satisfies ${\alpha }_k^I|0;p⟩=0$ for $k>0$ and $b_r^I|0;p⟩=0$ for $r>0$. A negative fermion mode is a fermionic creation operator for a transverse worldsheet excitation. The Neveu–Schwarz level operator and mass condition areThe smallest positive frequency is $1/2$, so the only first-excited states areThe same vector-polarization argument requires them to be massless in Lorentz-consistent quantization, fixing $a_{\mathrm{NS}}=1/2$.

At $N=1$, ${\alpha }_{-1}^I|0;p⟩$ is a vector, while $b_{-1/2}^I{b}_{-1/2}^J|0;p⟩$ is the exterior square: interchanging the indices changes the sign and equal indices give zero. Together they branch from an antisymmetric tensor:Thus the Neveu–Schwarz level-one massive tensor has $(D-1)(D-2)/2$ polarizations and mass squared $1/(2{\alpha }^{\prime })$. It differs from spin two because the two-fermion tensor is antisymmetric and has neither the symmetric trace-free representation nor its scalar trace. At $D=10$, the count is $8+28=36$, compared with $44$ for a massive spin-two field. The specified states are before the GSO projection; the usual tachyon-removing GSO projection also removes this integer level.

Use the mostly-plus Minkowski metric, with the last spatial coordinate singled out:Then $2d{x}^{+}d{x}^{-}=(d{x}^{D-1}{)}^2-(d{x}^0{)}^2$, as required. In these light-cone coordinates, ${\partial }^{+}={\partial }_{-}$, ${\partial }^{-}={\partial }_{+}$ and ${\square }_D=2{\partial }_{+}{\partial }_{-}+{\partial }_I{\partial }_I$.

Introduce the Kalb-Ramond field strengthThe field equation is ${\partial }^p{H}_{pmn}=0$. Under $\delta B_{mn}={\partial }_m{\alpha }_n-{\partial }_n{\alpha }_m$, every second-derivative contribution to $\delta H$ cancels by commutation of derivatives. Thus $H$ and its equation are gauge-invariant. This is the differential-form identity $d^2=0$ for the two-form gauge field $B$.

To impose light-cone gauge for a two-form, first choose ${\alpha }_{-}=0$ andThis sets $B_{-m}$ to zero. A residual transformation preserving the gauge obeys ${\partial }_{-}{\alpha }_m={\partial }_m{\alpha }_{-}$. With $\chi ={\partial }_{-}^{-1}{\alpha }_{-}$, this implies ${\alpha }_m={\partial }_m\chi$, including $m=-$. Such a parameter changes $B$ by zero. Therefore no nontrivial gauge transformation of $B$ remains in the sector where ${\partial }_{-}$ is invertible. There is still a redundant description of the gauge parameter itself, $\alpha \mapsto \alpha +d\chi$; the excluded ${\partial }_{-}$ zero modes would require separate treatment.

The $(-,n)$ field equations now say ${\partial }_{-}{\partial }^p{B}_{np}=0$, hence ${\partial }^p{B}_{np}=0$. For $n=I$ this givesFor $n=+$, ${\partial }^I{B}_{+I}=0$ follows automatically from antisymmetry of $B_{IJ}$. Therefore the independent components and their equation areThe remaining equations follow from this wave equation and the reconstructed longitudinal components.

At the first massless level of the closed bosonic string, states ${\alpha }_{-1}^I{\tilde{\alpha }}_{-1}^J|p⟩$ carry a product of two transverse vector polarizations. Its symmetric traceless, antisymmetric and trace parts are respectively the graviton, the Kalb–Ramond field and the dilaton. The antisymmetric part has exactly the polarization count just found.

The gauge-invariant string coupling to a two-form uses the pullback of the background Kalb–Ramond field to the string worldsheet:Its variation is $q{\int }_{\Sigma }d(X^{\ast }\alpha )=q{\int }_{\partial \Sigma }{X}^{\ast }\alpha$, by Stokes theorem. It vanishes on a closed worldsheet. For a cylindrical propagation surface it is only an initial/final boundary term, handled by fixed-boundary gauge parameters or the corresponding transformation of external states. Thus the closed string is naturally charged under a two-form gauge field.

A continuous Lorentz transformation in one spatial dimension must preserve $t^2-x^2$. In the component connected to the identity, take the image of the time axis to have velocity $v$, with $|v|<1$, and write $B_v(1,0{)}^{\mathsf{T}}=\gamma (1,v{)}^{\mathsf{T}}$. Preservation of the Minkowski norm gives ${\gamma }^2(1-v^2)=1$, so $\gamma =(1-v^2{)}^{-1/2}$. The spacelike column must be orthogonal to the timelike column and have squared Minkowski norm $-1$; continuity fixes it to $\gamma (v,1{)}^{\mathsf{T}}$. ThusHere $x=(t,x{)}^{\mathsf{T}}$ denotes the coordinate pair. This is the positive off-diagonal convention in the source; describing a passive frame moving at $+v$ instead reverses the sign of $v$.

Define the rapidity $\alpha =\mathrm{artanh}v$. The hyperbolic function identities giveIn light-cone coordinates, $t^{\prime }+x^{\prime }=e^{\alpha }(t+x)$ and $t^{\prime }-x^{\prime }=e^{-\alpha }(t-x)$. Therefore the derivatives transform oppositely:Use the field-transformation convention ${\psi }^{\prime }(x^{\prime })=S(\alpha )\psi (x)$. The component action in c is invariant underIndeed, the right-component kinetic term gains $e^{\alpha /2}{e}^{-\alpha }{e}^{\alpha /2}=1$, and the left-component kinetic term gains $e^{-\alpha /2}{e}^{\alpha }{e}^{-\alpha /2}=1$. The two factors in each Dirac mass term cancel as well. Finally $\det B=1$, so $d{t}^{\prime }d{x}^{\prime }=dtdx$ and the whole action is unchanged. These are the chiral boost weights in two spacetime dimensions.

In spinor notation, $S(\alpha )=\exp (\alpha {\gamma }^5/2)$ satisfies $S^{-1}{\gamma }^{\mu }S=B^{\mu }{}_{\nu }{\gamma }^{\nu }$ and $S^{†}{\gamma }^{0}S={\gamma }^0$, so ${\bar{\psi }}^{\prime }=\bar{\psi }{S}^{-1}$. These identities directly verify covariance of the Dirac equation and the Lorentz invariance of its action. The matrices $S$ are not unitary with respect to the ordinary component norm; the Dirac adjoint supplies the invariant bilinear.

The statement about there being only Lorentz boosts applies to the identity component of the One-dimensional Lorentz group. The full Lorentz group also has discrete transformations: the concrete spatial reflection $\mathrm{diag}(1,-1)$ preserves the Minkowski metric but has determinant $-1$, so it cannot equal any displayed $B(\alpha )$, whose determinant is $+1$.